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8
7
6
5
4
3
2
1
By Eq. (24.5),
Vb ��2 mV
tp (6 � 2) ms 4
Duty cycle � � ��� ���0.4 (decimal form)
T 10 ms 10
Peak value (from 0-V reference) � 8 mV
Vav � (duty cycle)(peak value) � (1 � duty cycle)(Vb) � (0.4)(8 mV) � (1 � 0.4)(2 mV)
� 3.2 mV � 1.2 mV � 4.4 mV
as obtained above.
EXAMPLE 24.7 Given a periodic pulse waveform with a duty cycle
of 28%, a peak value of 7 V, and a base-line voltage of �3 V:
a. Determine the average value.
b. Sketch the waveform.
c. Verify the result of part (a) using the method of Section 13.6.
Solutions:
a. By Eq. (24.5),
Vav � (duty cycle)(peak value) � (1 � duty cycle)(Vb) � (0.28)(7 V) � (1 � 0.28)(�3 V) � 1.96 V � (�2.16 V)
� �0.2 V
b. See Fig. 24.16.
(7 V)(0.28T) � (3 V)(0.72T)
c. G ���� � 1.96 V � 2.16 V
T
� �0.2 V
as obtained above.
Instrumentation
0
v (mV)
5 10 15 20 t (�s) �
T
FIG. 24.15
Example 24.6.
The average value (dc value) of any waveform can be easily determined
using the oscilloscope. If the mode switch of the scope is set in the ac
position, the average or dc component of the applied waveform will be
blocked by an internal capacitor from reaching the screen. The pattern
can be adjusted to establish the display of Fig. 24.17(a). If the mode
switch is then placed in the dc position, the vertical shift (positive or
V av
7 V
0
–3 V
v
0.28T
AVERAGE VALUE ⏐⏐⏐ 1101
T
FIG. 24.16
Solution to part (b) of Example 24.7.
div.
div.
0
0
v
v
Vertical sensitivity = 5 mV/div.
Horizontal sensitivity = 5 �s/div. �
(a)
ac mode
Vertical sensitivity = 5 mV/div.
Horizontal sensitivity = 5 �s/div. �
(b)
Vav = 4 mV
t
dc mode
FIG. 24.17
Determining the average value of a pulse
waveform using an oscilloscope.
t
t